Problem: Let $\mathbf{a}$ and $\mathbf{b}$ be vectors such that $\|\mathbf{a}\| = 2,$ $\|\mathbf{b}\| = 5,$ and $\|\mathbf{a} \times \mathbf{b}\| = 8.$  Find $|\mathbf{a} \cdot \mathbf{b}|.$
Recall that
\[\|\mathbf{a} \times \mathbf{b}\| = \|\mathbf{a}\| \|\mathbf{b}\| \sin \theta,\]where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}.$  Hence,
\[8 = 2 \cdot 5 \cdot \sin \theta,\]so $\sin \theta = \frac{4}{5}.$  Then
\[\cos^2 \theta = 1 - \sin^2 \theta = \frac{9}{25},\]so $\cos \theta = \pm \frac{3}{5}.$  Hence,
\[|\mathbf{a} \cdot \mathbf{b}| = \|\mathbf{a}\| \|\mathbf{b}\| |\cos \theta| = 2 \cdot 5 \cdot \frac{3}{5} = \boxed{6}.\]